Question
The $$x - t$$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $$t = \frac{4}{3}s$$ is
The $$x - t$$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $$t = \frac{4}{3}s$$ is
A.
$$\frac{{\sqrt 3 }}{{32}}{p^2}cm/{s^2}$$
B.
$$\frac{{ - {\pi ^2}}}{{32}}cm/{s^2}$$
C.
$$\frac{{{\pi ^2}}}{{32}}cm/{s^2}$$
D.
$$ - \frac{{\sqrt 3 }}{{32}}{\pi ^2}cm/{s^2}$$
Answer :
$$ - \frac{{\sqrt 3 }}{{32}}{\pi ^2}cm/{s^2}$$
Solution :
From the graph it is clear that the amplitude is $$1cm$$ and the time period is 8 second. Therefore the equation for the S.H.M. is $$x = a\sin \left( {\frac{{2\pi }}{T}} \right) \times t = 1\sin \left( {\frac{{2\pi }}{8}} \right)t = \sin \frac{\pi }{4}t$$
The velocity $$\left( v \right)$$ of the particle at any instant of time $$'t'$$ is $$v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sin \left( {\frac{\pi }{4}} \right)t} \right] = \frac{\pi }{4}\cos \left( {\frac{\pi }{4}} \right)t$$
The acceleration of the particle is $$\frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \left( {\frac{\pi }{4}} \right)t$$
At $$t = \frac{4}{3}s$$ we get
$$\eqalign{ & \frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \frac{\pi }{4} \times \frac{4}{3} = \frac{{ - {\pi ^2}}}{{16}}\sin \frac{\pi }{3} \cr & = \frac{{ - \sqrt 3 {\pi ^2}}}{{32}}cm/{s^2} \cr} $$
From the graph it is clear that the amplitude is $$1cm$$ and the time period is 8 second. Therefore the equation for the S.H.M. is $$x = a\sin \left( {\frac{{2\pi }}{T}} \right) \times t = 1\sin \left( {\frac{{2\pi }}{8}} \right)t = \sin \frac{\pi }{4}t$$
The velocity $$\left( v \right)$$ of the particle at any instant of time $$'t'$$ is $$v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sin \left( {\frac{\pi }{4}} \right)t} \right] = \frac{\pi }{4}\cos \left( {\frac{\pi }{4}} \right)t$$
The acceleration of the particle is $$\frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \left( {\frac{\pi }{4}} \right)t$$
At $$t = \frac{4}{3}s$$ we get
$$\eqalign{ & \frac{{{d^2}x}}{{d{t^2}}} = - {\left( {\frac{\pi }{4}} \right)^2}\sin \frac{\pi }{4} \times \frac{4}{3} = \frac{{ - {\pi ^2}}}{{16}}\sin \frac{\pi }{3} \cr & = \frac{{ - \sqrt 3 {\pi ^2}}}{{32}}cm/{s^2} \cr} $$